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Copyright N°. 



COFVRIGHT DEPOSrr 



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ANALYTICAL MECHANICS 



FOR STUDENTS OF PHYSICS AND 
ENGINEERING 



Call* 






BY 



HAKOUTUNE M, DADOUEIAN, M.A., Ph.D. 

Instructor of Physics in the Sheffield Scientific School 
of Yale University 




NEW YOEK 
D. VAN NOSTRAND COMPANY 

25 Park Place 
1913 



Di4 



Copyright, 1913, 

BY 

D. VAN NOSTRAND COMPANY 



Stanbopc fl>res& 

F H. GILSON COMPANY 
30ST0N, U.S. A 



©CI.A34301 



ANALYTICAL MECHANICS 



INTRODUCTION. 

1. Scope and Aim of Mechanics. Mechanics is the science 
of motion. It has a twofold object: 

First, to describe the motions of bodies and to interpret 
them by means of a few laws and principles, which are gen- 
eralizations derived from observation and experience. 

Second, to predict the motion of bodies for all times when 
the circumstances of the motion for any one instant are 
given, in addition to the special laws which govern the 
motion. 

The present tendency in science is toward regarding all 
physical phenomena as manifestations of motion. Compli- 
cated and apparently dissimilar phenomena are being ex- 
plained by the interactions and motions of electrons, atoms, 
molecules, cells, and other particles. The kinetic theory of 
heat, the wave theories of sound and light, and the electron 
theory of electricity are examples which illustrate the tend- 
ency toward a mechanical interpretation of the physical 
universe. 

This tendency not only emphasizes the fundamental im- 
portance of the science of mechanics to other physical 
sciences and engineering but it also broadens the aim of the 
science and makes the dynamical interpretation of all physi- 
cal phenomena its ultimate object. The aim of elementary 
mechanics is, however, very modest and its scope is limited to 
the discussion of the simplest cases of motion and equilibrium 
which occur in nature. 

l 



2 ANALYTICAL MECHANICS 

2. Divisions of Mechanics. It is customary to divide 
Mechanics into Kinematics and Dynamics. The former 
treats of the time and space relations of the motions of 
bodies withdtit regard to the interactions which cause them. 
In other words, Kinematics is the geometry of motion. In 
Dynamics, on the other hand, motion and equilibrium are 
treated as the results of interactions between bodies; conse- 
quently not only time and space enter into dynamical discus- 
sions, but also mass, the third element of motion. 

Dynamics in its turn is divided into Statics and Kinetics. 
Statics is the mechanics of bodies in equilibrium, while 
Kinetics is the mechanics of bodies in motion. 

Chapters II, III, and IV of the present work are devoted 
to problems in statics, while the rest of the book with the 
exception of Chapters I, V, and VII, are given to discussions 
of problems in kinetics. The subject matter of Chapters I 
and VII is essentially of a mathematical nature. In the 
former the addition and resolution of vectors are discussed, 
while in the latter the Calculus is applied to finding centers 
of mass and moments of inertia. Chapter V is devoted 
mainly to kinematic al problems. 



Q 



CHAPTER I. 
ADDITION AND RESOLUTION OF VECTORS. 

3. Scalar and Vector Magnitudes. — Physical magnitudes may 
be divided into two classes according to whether they have 
the property of orientation or not. Magnitudes which 
have direction are called vectors, while those which do not 
have this property are called scalars. Displacement, veloc- 
ity, acceleration, force, torque, and momentum are vector 
magnitudes. Mass, density, work, energy, and time are 
scalars. 

4. Graphical Representation of Vectors. — Vectors are repre- 
sented by directed lines or arrows. The length of the 
directed line represents the magnitude 
of the vector, while its direction coin- 
cides with that of the vector. For 
brevity the directed lines as well as 
the physical quantities which they 
represent are called vectors. The 
head and the tail of the directed line 
are called, respectively, the terminus 

and the origin of the vector. In Fig. 1, for instance, P is 
the origin and Q the terminus of the vector a. 

5. Notation. — Vectors will be denoted by letters printed in 
Gothic type, while their magnitudes will be represented by 
the same letters printed in italic type. Thus in Fig. 1 the 
vector PQ is denoted by a, but if it is desired to represent 
the length PQ without regard to its orientation a is used. 

6. Equal Vectors. — Two vectors are said to be equal if they 
have the same length and the same direction. It follows, 
therefore, that the value of a vector is not changed when it is 

3 



P 



4 ANALYTICAL MECHANICS 

moved about without changing its direction and magni- 
tude. 

7. Addition of Two Vectors. — Let the vectors a and b, Fig. 2, 
represent two displacements, then their sum is another 
vector, c, which is equivalent to the given vectors. In 
order to find c let us apply to a particle the operations indi- 
cated by a and b. Each vector displaces the particle along 
its direction through a distance equal to its length. There- 





Fig. 2. : > 

fore applying a to the particle at P, Fig. 3, the particle is 
brought to the point Q. Then applying the operation indi- 
cated by b the particle is brought to the point R. There- 
fore the result of the two operations is a displacement from 
P to R. But this is equivalent to a single operation repre- 
sented by the vector c, which has P for its origin and R for 
its terminus. Therefore c is called the sum, or the resultant, 
of a and b. This fact is denoted by the following vector 
equation, 

a + b = c. (I) 

8. Order of Addition. — The order of addition does not affect 
the result. If in Fig. 3 the order of the operations indicated 
by a and b is reversed the particle moves from P to Q' and 
then to R. Thus the path of the particle is changed but not 
the resultant displacement. 

9. Simultaneous Operation of Two Vectors. — The operations 
indicated by a and b may be performed simultaneously 
without affecting the final result. In order to illustrate 



ADDITION AND RESOLUTION OF VECTORS 5 

the simultaneous operation of two vectors suppose the 
particle to be a bead on the wire AB, Fig. 4. Move the 
wire, keeping it parallel to itself, until each of its particles 
is given a displacement represented by b. Simultaneously 
with the motion of the wire move the bead along the 
wire giving it a displacement equal to a. At the end of 









R 






-^ 


+f 


b/ 


>' 


*>? 


— r— — 
/ 




^f 

**■' 






(<■ 




/ 






a 


" M_ B 





Fig. 4. 

these operations the bead arrives at the point R. If both 
the wire and the bead are moved at constant rates the 
resultant vector c represents not only the resulting dis- 
placement but also the path of the particle. 

10. Rules for Adding Two Vectors. — The results of the last 
three paragraphs furnish us with the following methods for 
adding two vectors graphically. 

Triangle Method. — Move one of the vectors, ivithout changing 
its direction, until its origin falls upon the terminus of the 
other vector, then complete the triangle by drawing a vector the 
origin of which coincides with that of the first vector. The new 
vector is the resultant of the given vectors. 

Parallelogram Method. — Move one of the vectors until its 
origin falls on that of the other vector, complete the parallelo- 
gram, and then draw a vector which has the common origin of 
the given vectors for its origin and which forms a diagonal of 
the parallelogram. The new vector is the resultant of the given 
vectors. 

11. Analytical Expression for the Resultant of Two Vectors. 
— Let a and b, Fig. 5, be two vectors and c their result- 



6 



ANALYTICAL MECHANICS 



ant. Then 3 solving the triangle formed by these vectors, 
we obtain 



and 



c 2 = a 2 +b 2 +2abcos<t> 
b sin <f> 



tan 6 = 



a-\-b cos 



(ii) 
an) 



where a, b, and c are the magnitudes of a, b, and c,' respec- 
tively, while <f> and 6 are the angles b and c make with a. 
Equation (II) determines the magnitude and equation (III) 
the direction of c. 




Fig. 5. 



Fig. 6. 



Special Cases, (a). If a and b have the same direction, 
as in Fig. 6a, then = 0. Therefore 



and 



c 2 = a 2 +b 2 +2ab, 
tan (9 = 0, 



c = a + 6, 
= 0. 



Thus c has the same direction as a and b, while its magni- 
tude equals the arithmetical sum of their magnitudes. 

(b). When a and b are oppositely directed, as in Fig. 6b, 
= 7T. Therefore 



c 2 = a 2 +b' 



and 



tan = 0, 



c = a — b, 
6 = 0. 



ADDITION AND RESOLUTION OF VECTORS 7 

Thus the magnitude of c equals the algebraic sum of the 
magnitudes of a and b, while its direction is the same as 
that of the larger of the two. It is evident that if the 
magnitudes of a and b are equal c vanishes. Therefore 
two vectors of equal magnitude and opposite directions are 
the negatives of each other. In other words, when the direc- 
tion of a vector is reversed its sign is changed. 

(c). When a and b are at right angles to each other, as in 



Fig. 6c, <t> = ~- Therefore 



c 2 = a*+b' 
b 



and tan = 

a 



12. Difference of Two Vectors. — Subtraction is equivalent to 
the addition of a negative quantity. Therefore, to subtract 
b from a we add — b to a. Thus 
we have the following rule for 
subtracting one vector from an- 
other. 

In order to subtract one vector 
from another reverse the one to be 
subtracted and add it to the other 
vector. 

It is evident from Fig. 7 that 
the sum and the difference of two vectors form the diagonals 
of the parallelogram determined by them. 

ILLUSTRATIVE EXAMPLES. 

A particle is displaced 10 cm. N. 30° E., then 10 cm. E. Find the 
resulting displacement. 

Representing the displacements and their resultant by the vectors a, 
b, and c, Fig. 8, we obtain 




ANALYTICAL MECHANICS 



c 2 = o 2 -f- b 2 + 2 ab cos 

= (10cm.) 2 +(10cm.) 2 +2xl0cm.Xl0cm.cos(60°) 

= 300 cm. 2 
c = 10 \/3 cm. 

= 17.3 cm.* 
6 sin0 



tan0 



a-\-b cos </> 

10 cm. sin (60°) 
10 cm. + 10 cm. cos (60°) 
= |V3. 
= 30°. , « 



Therefore the resultant displacement 

is about 17.3 cm. along the direction N. 60° E. 




Fig. 8. 



PROBLEMS. 

1. A vector which points towards the East has a length of 16 cm., and 
another vector which points towards the Southeast is 25 cm. long. Find 
the direction and the magnitude of their sum. 

2. Find the direction and the magnitude of the difference of the 
vectors of the last problem. 

3. The sum of two vectors is perpendicular to their difference. Show 
that the vectors are equal in magnitude. 

4. The sum and the difference of two vectors are equal. Show that 
the vectors are at right angles to each other. 



13. Resolution of Vectors into Compo- 
nents. — The projection of a vector upon a 
line is called the component of the vector 
along that line. The vectors a* and a y in 
Fig. 9, for instance, are the components of 
a along the rr-axis and the 2/-axis, respec- 
tively. The following relations are evident 
from the figure and do not need further 
explanation. 




Fig. 9. 



* The symbol " = " will be used to denote approximate equality. There- 
fore " = " should be read " equals approximately," or "equals about," or 
"equals nearly." See p. 



ADDITION AND RESOLUTION OF VECTORS 



a — Six i" 3-j/? 

a x = a cos 0, 



«7V 



tan = 



(IV) 

(V) 

(VI) 

(VII) 



When a has components along all three axes of a rectangular 
system, Fig. 10, the following equations express the vector 
in terms of its components. 
Y 




Fig. 10. 

a= a z + a y + a 3 . (IV) 

a x = a cos a\ 

a y = a cos a 2 (V) 

a z = a cos a 3 

a = Va^+V + a, 2 , (VI') 

where en, a 2 , and « 3 are the angles a makes with the coordi- 
nate axes. 

14. Resultant of Any Number of Vectors. Graphical Methods. 
— The resultant of a number of vectors a, b, c, etc., may be 
obtained by either of the following methods. 

First : move b, without changing either its direction or its 
magnitude, until its origin falls on the terminus of a, then 



10 



ANALYTICAL MECHANICS 




Fig. 11. 



move c until its origin falls on the terminus of b, and so on 
until all the vectors are joined. This gives, in general, an 
open polygon. Then the resultant 
is obtained by drawing a vector 
which closes the polygon and which 
has its origin at the origin of a. 
The validity of this method will be 
seen from Fig. 11, where r repre- 
sents the resultant vector. Evi- 
dently the resultant vanishes when 
the given vectors form a closed 
polygon. 

Second : draw a system of rectangular coordinate axes ; 
resolve each vector into components along the axes; add the 
components along each axis geometrically, beginning at 
the origin. This gives the components of the required vector. 
Then draw the rectangular parallelopiped determined by 
these components. The resultant is a vector which has the 
origin of the axes for its origin and forms a diagonal of the 
parallelopiped. This method is based upon the following 
analytical method. 

15. Analytical Method. — Expressing the given vectors and their 
resultant in terms of their rectangular components, we have 

a = 3.x ~r &y t~ a 2 , 
b = b x + b y + b z , 



(1) 



r = r x + x y + r 2 
Substituting from (1) in the vector equation 

r=>+b+c+ • • • A (2) 

and collecting the terms we obtain 

r*+> y +r 2 = (a x +[b x +}]^0+(av+ b,,+ • ■ •) 

+ (a.+ b. + ; ■ •)• (3) 

But since the directions of the coordinate axes are indepen- 
dent, the components of r along any one of the axes must 



ADDITION AND RESOLUTION OF VECTORS 



11 



equal the sum of the corresponding components of the given 
vectors. Therefore (3) can be split into the following three 
separate equations. 

r*= a x + b x + c x + • • • , 

r y = n y +b y +c y + • • • , (4) 

x z = a 2 + b z + c z + • • • . . 

It was shown in § 11 that when two vectors are parallel 
the algebraic sum of their magnitudes equals the magni- 




Fig. 12. 



tude of their resultant. This result may be extended to 
any number of parallel vectors. Therefore we can put the 
vector equations of (4) into the following algebraic forms. 

r x = a x +b x + c x + - • • ,' 

r y = a y + b y + c y + - • - , ■ (5) 

r z = a z + b z + c z + • • • . 

Equations (5) determine r through the following relations 

r = a//\ 






COS ai = — : 

r 



COS a 2 



— , COS a z = — : 

r r 



(6) 

(7) 



where a h a 2 , and a 3 are the angles r makes with the axes. 

16. Multiplication and Division of a Vector by a Scalar. — 
When a vector is multiplied or divided by a scalar the result 
is a vector which has the same direction as the original vec- 
tor. If, in the equation b = ma, m be a scalar then b has the 
same direction as a but its magnitude is m times that of a. 



12 



ANALYTICAL MECHANICS 



ILLUSTRATIVE EXAMPLE. 

A man walks 3 miles N. 30° E., then one mile E., then 3 miles S. 45° E., 
then 4 miles S., then one mile N. 30° W. Find his final position. 

Representing the displacements by vectors we obtain the graphical 
solution given in Fig. 13, where r represents the resultant displacement. 




Fig. 13. 



In order to find r analytically we first determine its components. Thus 

r x = [3 cos (60°) + cos (0°) + 2 cos (-45°) + 4 cos (-90°) 
+ cos (120°)] miles 
= (2 + V2) miles 
= 3.41 miles. 
r y = [3 sin (60°) + sin (0°) + 2 sin (-45°) + 4 sin (-90°) 
+ sin (120°)] miles 
= (2 a/3 -a/2 -4) miles 
= —1.95 miles. 



r = Vr x 2 + r y 2 
= 3.93 miles. 



The direction of r is given by the following relation. 

a r„.. -1.95 

w = r, = iur' 

... e = -37M. 

Therefore the final position of the man is about 3.93 miles S. 52°.9 E. 
from his starting point. 



ADDITION AND RESOLUTION OF VECTORS 13 

PROBLEMS. 

1. The resultant of two vectors which are at right angles to each other 
is twice the smaller of the two. The magnitude of the smaller vector is 
a; find the magnitude of the greater vector. 

2. In the preceding problem find the resultant vector. 

3. Find analytically the sum of three equal vectors which point in the 
following directions — East, N. 30° W., and S. 30° W. 

4. In the preceding problem make use of the first graphical method. 

5. In problem 3 make use of the second graphical method. 

6. A vector which is 15 cm. long points N. 30° E. Find its compo- 
nents in the following directions. 

(a) N. 30° W. (c) W. (e) S. 60° E. 

■(b) N. 60° E. (d) S. 30° W. (f) E. 

7. A vector a is in the xy-p\a,ne. If 3 is added to a x and 4 to a y the 
magnitude of the new vector equals a x -f a y . Find the magnitude and 
direction of a. 

8. Three vectors a, b, and c lie in the :n/-plane. Find their resultants 
analytically, taking the magnitudes of their components from the follow- 
ing tables: 





a x 


a y 


b x 


by 


C x 


Cy 


(1) 


6, 


9, 


-5, 


2, 


o, 


10 


(2) 


-3, 


7, 


5, 


o, 


6, 


-8 


(3) 


o, 


-10, 


8, 


5, 


3, 


-2 


(4) 


2, 


o, 


-6, 


4, 


o, 


8 



9. In the preceding problem make use of the second graphical method. 

10. Straight horizontal tunnels in a mine connect the points Pi, P 2 , 
P 3 , and Pi, in the given order. The length of each tunnel and the angle 
it makes with the meridian are given in the following tables. Find the 
lengths and directions of the tunnels which have to be dug in order to 
connect Pi with P 3 and P 4 . 

PiP 2 = 200 feet, and makes 30° with the meridian. 
P 2 P 3 = 100 feet, and makes 120° with the meridian. 
P3P4 = 400 feet, and makes 300° with the meridian. 

11. Work out the preceding problem by the first graphical method. 

12. Work out problem 10 by the second graphical method. 

13. Find the direction and magnitude of the force experienced by an 
electrical charge of five units placed at one vertex of an equilateral tri- 
angle due to two unlike charges of 10 units each placed at the other vertices. 
The sides of the triangle are 2 cm. 



CHAPTER II. 
EQUILIBRIUM OF A PARTICLE. 

ACTION AND REACTION. FORCE. 

17. Particle. — A body whose dimensions are negligible is 
called a particle. In a problem any body may be considered 
as a particle so long as it does not tend to rotate. Even 
when the body rotates it may be considered as a particle if 
its rotation does not enter into the problem. For instance, 
in discussing the motion of the earth in its orbit the earth 
is considered as a particle, because its rotation about its 
axis does not enter into the discussion. 

18. Degrees of Freedom. — The number of independent ways 
in which a body can move is called the number of degrees of 
freedom of its motion. It equals the number of coordinates 
which are necessary in order to specify completely the posi- 
tion of the body. A free particle can move in three inde- 
pendent directions, that is, along the three axes of a system 
of rectangular coordinates, therefore it has three degrees of 
freedom. When the particle is constrained to move in a 
plane its freedom is reduced to two degrees, because it can 
move only in two independent directions. When it is con- 
strained to move in a straight line it has only one degree of 
freedom. 

19. Force. — While considering the motion or the equilibrium 
of a body our attention is claimed not only by that body 
but also by others which act upon it. In order to insure 
concentration of attention problems in Dynamics are sim- 
plified in the following manner. All bodies are eliminated, 
except the one the motion of which is being discussed, and 
their actions upon the latter are represented by certain vec- 
tor magnitudes known as forces. As an illustration consider 

14 



l 




EQUILIBRIUM OF A PARTICLE 15 

the equilibrium of the shaded part of the rope in Fig. 14a. 
The shaded part is acted upon 

by the adjoining sections of the "^ S3 i) 

rope. Therefore we consider the 
shaded part alone and represent 
the actions of the adjoining parts 
by the forces F and F', as shown 
in Fig. 14b. 

20. Definition of Force. — Force is a vector magnitude which 
represents the action of one body upon another. The interac- 
tion between two bodies takes place across an area, while the 
forces which represent them are supposed to be applied at 
one point. Therefore the introduction of the idea of force 
presupposes the simplification of dynamical problems which 
is obtained by considering bodies as single particles, or as 
a system of particles. 

21. Internal Force. — A force which represents the action of 
one part of a body upon another part of the same body is 
called an internal force. 

22. External Force. — A force which represents the action of 
one body upon another body is called an external force. 

23. Unit Force. — The engineering unit of force among 
English speaking people is the pound. The pound is the 
weight, in London, of a certain piece of platinum kept by 
the British government. 

24. The Law of Action and Reaction. — The fundamental law 
of Mechanics is known as the law of action and reaction. 
Newton (1692-1727), who was the first to formulate it, put 
the law in the following form. 

"To every action there is an equal and opposite reaction, 
or the mutual actions of two bodies are equal and oppositely 
directed." 

Let us apply this law to the interaction between a book 
and the hand in which you hold it. Your hand presses 
upward upon the book in order to keep it from falling, 



16 ANALYTICAL MECHANICS 

while the book presses downward upon your hand. The 
law states that the action of your hand equals the reaction 
of the book and is in the opposite direction. The book 
reacts upon your hand because the earth attracts it. When 
your hand and the earth are the only bodies which act upon 
the book, the action of your hand equals and is opposite to 
the action of the earth. In other words the sum of the 
two actions is nil. Generalizing from this simple illustra- 
tion we can put the law into the following form: 

To every action there is an equal and opposite reaction, or 
the sum of all the actions to which a body or a part of a body 
is subject at any instant vanishes : 

SA = 0. (A) 

25. Condition for the Equilibrium of a Particle. — The condi- 
tion of equilibrium of a particle is obtained by replacing the 
term " action" by the term "force" in the last form of the 
fundamental law and then stating it in the form of a condi- 
tion. Thus — in order that a particle be in equilibrium the 
sum of all the forces which act upon it must vanish. 

In other words if Fi, F 2 , F 3 , . . ., F n are the forces which 
act upon a particle, then the vector equation 

Fi+F 2 +F 3 + ■"■ ■ +F.= (I) 

must be satisfied in order that the particle be in equilib- 
rium. Equation (I) is equivalent to stating that when the 
forces are added graphically they form a closed polygon. 
But when the sum of a number of vectors vanishes the sum 
of their components also vanishes. Therefore we must 
have 

x x +x 2 + • .-*. +X n = 0,1 

Yi+Y 2 + • • • +Y n =0, (no 

Zi + Z 2 + • • • + Z n = 0, J 

where X;, Y», and Z z are the components of F t -.* Since the 

* The subscript "»" is used to denote "any one," thus Fj denotes any one 
of Fi, F 2 , etc. 



EQUILIBRIUM OF A PARTICLE 17 

vectors in each of the equations of (IF) are parallel we can 
write them as algebraic equations. Therefore we have the 
following equations for the analytical form of the condition 
of equilibrium of a particle. 



2Z = Xi + X 2 + • • • +X n =0,*l 



(II) 



2F = Yl + Y 2 + • • • + Y n = 0, 
2Z eee Z x + Z 2 + • • • + Z n = 0. 

The condition of equilibrium may, therefore, be stated 
in the following form. 

In order that a particle be in equilibrium the algebraic sum 
of the components of the forces along each of the axes of a rec- 
tangular system of coordinates must vanish. 

The following rules will be helpful in working out prob- 
lems on the equilibrium of a particle. 

First. Represent the particle by a point and the action of 

each body which acts upon it by a properly chosen 

force-vector. Be sure that all the bodies which 

act upon the particle are thus represented. 

Second. Set the sums of the components of the forces 

along properly chosen axes equal to zero. 
Third. If there are not equations enough to determine the 
unknown quantities, obtain others from the geo- 
metrical connections of the problem. 
Fourth. Solve these equations for the required quantities. 
Fifth. Discuss the results. 

ILLUSTRATIVE EXAMPLES. 

1. A particle suspended by a string is pulled aside by a horizontal 
force until the string makes an angle a with the vertical. Find the tensile 
force in the string and the magnitude of the horizontal force in terms of 
the weight of the particle. 

The particle is acted upon by three bodies, namely, the earth, the 
string, and the body which exerts the horizontal force. Therefore, we 

* The relation 2X = Xi + X 2 + • • • + X n is not an equation. It 
merely states that ZX is identical with and is an abbreviation for Xy + 
X« + • • • + x n . 



18 



ANALYTICAL MECHANICS 



represent the actions of these bodies by three force- vectors, W, T, and F, 
Fig. 15, and then apply the conditions of equilibrium. Setting equal to 
zero the sums of the components of the y 
forces along the x- and ?/-axes, we get 

XX = F - T sin a = 0. (a) 

2F - -W+Tcosa = 0. (b) 

Solving equations (a) and (b) we have 
W 



and 



T = 

cos a 

F = T sin a 
= W tan a. 



Discussion. — When a = 0, T=W 



and F = 0. 



When a = - , T = oo and 

Z 




Fig. 15. 



F = oo. Therefore no finite horizontal 
force can make the string perfectly hori- 
zontal. 

2. A uniform bar, of weight W and 
length a, is suspended in a horizontal 
position by two strings of equal length 
I. The lower ends of the strings are fastened to the ends of the bar and 
the upper ends to a peg. Find the tensile force in the strings. 

The bar is acted upon by three bodies, namely the earth and the two 

strings. We represent their actions by the forces W, Ti, and T2, Fig. 16a. 

The tensile forces of the strings act at the ends of the bar. On the other 

hand the weight is distributed all along the rod. But we may consider 

it as acting at the middle point, as in Fig. 16a, or we may replace the rod 

W 
by two particles of weight — each, as shown in Fig. 16b. In the last case 

the rigidity of the bar which prevents its ends from coming together is 
represented by the forces F and — F. 

Considering each particle separately and setting equal to zero the sums 
of the components of the forces along the axes, we obtain 

2X ^ T x cos a - F = 0, 

W 
2F = 7 7 1 sino;-^ = 0, 

Zi 

for the first particle, and 

zZX m -T 2 cos a + F = 0, 

SF = 7 7 2 sina-^ = 0, 

z 



EQUILIBRIUM OF A PARTICLE 



19 



for the second particle. It follows from these equations that 



W 



2 sin a 
I 



ViP 



w. 




Fig. 16. 

Discussion. — The tensile force of the strings increases indefinitely 
as their total length approaches that of the bar. On the other hand 
as the length of the strings becomes very large compared with that of the 

W 

bar the tensile force approaches — as a limit. 

A 

The problem can be solved also by considering the forces acting on the 
peg, as shown in Fig. 16b. 



PROBLEMS. 

1. Show that when a particle is in equilibrium under the action of two 
forces, the forces must lie in the same straight line. 

2. Show that when a particle is in equilibrium under the action of 
three forces the forces lie in the same plane. 



20 ANALYTICAL MECHANICS 

3. Find the horizontal force which will keep in equilibrium a weight 
of 150 pounds on a smooth inclined plane which makes 60° with the 
horizon. 

4. A ring of weight W is suspended by means of a string of length I, 
the ends of which are attached to two points on the same horizontal line. 
Find the tensile force of the string if the distance between its ends is d. 
Also discuss the limiting cases in which I approaches d or becomes very 
large compared with it. 

5. A body of weight W is suspended by two strings of lengths k and h. 
The upper end of each string is attached to a fixed point in the same 
horizontal line. Find the tensile forces in the strings if the distance 
between the two points is d. 

6. A weight is suspended by four equal strings, the upper ends of which 
are attached to the vertices of a horizontal square. Find the tensile 
forces in the strings. 

7. A particle is in equilibrium on a smooth inclined plane under the 
action of two equal forces, the one acting along the plane upwards and 
the other horizontally. Find the inclination of the plane. 

8. Apply the conditions of equilibrium to find the magnitude and 
direction of the resultant of a number of forces acting upon a particle. 

9. Two spheres of equal radius and equal weight are in equilibrium in 
a smooth hemispherical bowl; find the reactions between the two spheres 
and between the spheres and the bowl. 

10. The ends of a string, 60 cm. long, are fastened to two points in the 
same horizontal line and at a distance of 

40 cm. apart; two weights are hung from 
points in the string 25 cm. and 20 cm. from 
the ends. Find the ratio of the weights if 
the part of the string between them is hori- 
zontal. 

11. A single triangular truss of 24 feet 
span and 5 feet depth supports a load 
of 3 tons at the apex. Find the forces acting on the rafters and the tie 
rod. 

12. A particle of weight W can be kept in equilibrium upon a smooth 
inclined plane by a force Fx acting horizontally; it can also be kept in 
equilibrium by a force F 2 acting parallel to the plane. Express W in 
terms of Fi and F 2 . 

13. In the following arrangements of pulleys find the relation between 
F and W. 




EQUILIBRIUM OF A PARTICLE 

L .j wwmmmm/A Y/////////////M 



21 





SLIDING FRICTION. 

26. Frictional Force. — Consider the forces acting upon a body 
which is in equilibrium on a rough inclined plane, Fig. 17. 
The body is acted upon by two 
forces, namely, its weight, W, 
and the reaction of the plane, 
R. The reaction of the plane 
is the result of two distinct 
and independent forces. One 
of these, N, is perpendicular 
to the plane and is called the 
normal reaction. The other, 
F, is along the plane and is 
called the frictional force. The 

normal reaction is due to the rigidity of the plane. It re- 
sists the tendency of the body to go through the plane. The 
frictional force is due to the roughness of the contact between 
the body and the plane. It prevents the body from sliding 
down the plane. 

27. Angle of Friction. — As we increase the angle of elevation 
of the inclined plane a certain definite angle will be reached 




Fig 



22 ANALYTICAL MECHANICS 

when the equilibrium is disturbed and the body begins to 
slide down the plane. This angle is called the angle of 
friction. This definition for the angle of friction does not 
hold when the body is acted upon by other forces besides 
its weight and the reaction of the plane. The following 
definition, however, is valid under all circumstances: The 
angle of friction equals the angle which the total reaction makes 
with the normal to the surface of contact when the body is on 
the point of motion. 

28. Coefficient of Friction. — Denoting the angle of friction 
by 0, we obtain 

F = R sin cf), 

N = R cos 0. 

Therefore F = N tan <f> 

= »N, (III) 

where n = tan and is called the coefficient of friction. 
The angle of friction and consequently the coefficient of 
friction are constants which depend upon the surfaces in 
contact. The last four equations hold true only when the body 
is on the point of motion. 

29. Static and Kinetic Friction. — The friction which comes 
into play is called static friction if the body is at rest and 
kinetic friction if it is in motion. 

30. Laws of Friction. — The following statements, which are 
generalizations derived from experimental results, bring out 
the important properties of friction. They hold true within 
certain limits and are only approximately true even within 
these limits. 

1. Frictional forces come into play only when a body is 
urged to move. 

2. Frictional forces always act in a direction opposite to 
that in which the body is urged to move. 

3. Frictional force is proportional to the normal reaction, 



EQUILIBRIUM OF A PARTICLE 



23 



4. Frictional force is independent of the area of contact. 

5. The static frictional force which comes into play is not 
greater than that which is necessary to keep the body in 
equilibrium. 

6. Kinetic friction is smaller than static friction. i 
Laws 1 to 4 hold true for both static and kinetic friction. 

The coefficient of friction between two bodies depends upon 
the condition of surfaces in contact. Therefore the value of 
ix is not a perfectly definite constant for a given pair of sub- 
stances in contact. 

The values given in the following table are averages of 
values obtained by several experimenters. 





Condition of surfaces in 
contact. 


Coefficient of friction. 




Static. 


Kinetic. 


Wood on wood 

Wood on wood 

Wood on wood 

Heavy rope on wood 

Heavy rope on wood 

Cast iron on cast iron 

Cast iron on cast iron 


Dry 

Wet 
Polished and greased 
Dry 
Wet 
Dry 

Greased 
Wet 


.50 
.68 
.35 
.60 
.80 
.24 
.15 
.65 
.30 


.36 
.25 

.12 
.40 
.35 

.18 
.13 


Cast iron on oak 

Leather on cast iron 





ILLUSTRATIVE EXAMPLES. 
1. A body which is on a rough horizontal floor can be brought to the 
point of motion by a force which makes an angle a with the floor. Find 
the reaction of the floor and the coefficient of friction. 
The body is acted upon by three forces, Fig. 18, 
P, the given force, 
W, the weight of the body, 
R, the reaction of the floor. 
Replacing R by its components F and N , and applying the conditions 
of equilibrium, we obtain 

XX = P cos a - F = 0, 
2Y = P sin a + N - W = 0. 
Therefore F = P cos a, 

JV = W - P sin a, 



24 



ANALYTICAL MECHANICS 



and 



R = VF 2 + JV 2 

= VP 2 + W 2 - 2 PW sin a. 



But since the body is on the point of 'motion the relation F = /jlN holds. 
Therefore 

_ _F _ P cos a 
M N W-P sin a' 



Discussion. — (a) When a = 0, 7 R = VP 2 + W 2 and ju = — ■ 
_ W 

(b) When a = ^, R = P-W = 0, therefore P = W, and n is indeter- 
minate, (c) When P _= 0, m = 0, and # = IT. 
Yl 





Fig. 19. 

2. A body which rests upon a rough inclined plane is brought to the 
point of motion up the inclined plane by a horizontal force. Find /* and R. 
The body is acted upon by three forces, Fig. 19, 
P, the horizontal force, 
W, the weight, 
R, the reaction of the plane. 

Replacing R by its components F and N, and taking the axes along and 

at right angles to the plane, we obtain 

SX s P cos a - F - W sin a = 0, 
27 = -P sin a + N - W cos a = 0. 



Therefore 



F = P cos a — W sin a, 
N = P sin a + W cos a, 

p = VfmT/v 2 
= Vp 2 + w 2 ~ 



and y 



EQUILIBRIUM OF A PARTICLE 25 

F P cos a — W sin a 



N P sin a + W cos a 
P 



Discussion. — (a) When a = 0, \x = — , and # = W V/jl 2 + 1. 

(b) When P = 0, ju = —tan a; therefore a = — <f>, that is, the inclined 
plane must be tipped in the opposite direction and must be given an 
angle of elevation equal to the angle of friction in order that motion may 
take place towards the positive direction of the a>axis. 

PROBLEMS. 

1. A body which weighs 100 pounds is barely started to move on a 
rough horizontal plane by a force of 150 pounds acting in a direction 
making 30° with the horizon. Find R and (jl. 

2. A body placed on a rough inclined plane barely starts to move 
when acted upon by a force equal to the weight of the body. Find 
the coefficient of friction, (a) when the force is normal to the plane; 
(b) when it is parallel to the plane. 

3. A horizontal force equal to the weight of the body has to be applied 
in order to just start a body into motion on a horizontal floor. Find the 
coefficient of friction. 

4. A weight W rests on a rough inclined plane, which makes an angle 
a with the horizon. Find the smallest force which will move the weight 
if the coefficient of friction is /jl. 

5. How would you determine experimentally the coefficient of friction 
between two bodies? 

6. A weight of 75 pounds rests on a rough horizontal floor. Find the 
magnitude of the least horizontal force which will move the body if the 
coefficient of friction is 0.4; also find the reaction of the plane. 

7. A particle of weight W is in equilibrium on an inclined plane under 
the action of a force F, which makes the magnitude of the normal pres- 
sure equal W. The coefficient of friction is /x and the angle of elevation of 
the inclined plane is a. Find the magnitude and direction of the force. 

8. An insect starts from the highest point of a sphere and crawls 
down. Where will it begin to slide if the coefficient of friction between 
the insect and the sphere is £? 

9. The greatest force, which can keep a particle at rest, acting along 
an inclined plane, equals twice the least force. Find the coefficient of 
friction. The angle of elevation of the plane is a. 

31. Resultant of a System of Forces. — The resultant of a 
number of forces which act upon a particle is a force which 



26 ANALYTICAL MECHANICS ' 

is equivalent to the given forces. There are two criteria by 
which this equivalence may be tested. First : The resultant 
force will give the particle the same motion, when applied 
to it, as that imparted by the given system of forces. We 
cannot use this test just now because we have not yet 
studied motion. Second: When the resultant force is re- 
versed and applied to the particle simultaneously with the 
given forces the particle remains in equilibrium. 

According to the second criterion, therefore, the resultant, 
R, of the forces Fi, F 2 , . . ., F„, must satisfy the equation 

-R+(F!+F 2 + • • • + F n ) = 0,l riv ,x 

or r = f 1 +F 2 + • • • +F n . J 

Splitting the last equation into three algebraic equations, 
we obtain 

x = x 1 +x 2 +- • . +z n , 

F = F 1 +y 2 +. • • +F n , (IV) 

z = z 1 + z 2 + . • • +z n , 

where X t , Y t , and Z { are the components of R;. 
The magnitude of R is given by the relation 

r = V X 2 +Y 2 +Z 2 , ( V) 

while the direction is obtained from the following expressions 
for its direction cosines. 

X Y Z 

COS a\ = — i COS a 2 = — > COS a 3 = -• (VI r ) 

k a k 

Special Case. — When the forces he in the xy-plane the 
^-components of each force equals zero. Therefore we have 

R = Vx 2 +Y 2 , __ (V) 

and tan0 = |i> (VI) 

X 

where is the angle R makes with the #-axis. 



EQUILIBRIUM OF A PARTICLE 27 

PROBLEMS. 

1. Three men pull on a ring. The first man pulls with a force of 50 
pounds toward the N. 30° W. The second man pulls toward the S. 45° 
E. with a force of 75 pounds, and the third man pulls with a force of 100 
pounds toward the west. Determine the magnitude and direction of 
the resultant force. 

2. Show that the resultant of two forces acting upon a particle lies 
in the plane of the given forces. 

3. Show that the line of action of the resultant of two forces lies 
within the angle made by the forces. 

4. Find the direction and magnitude of the resultant of three equal 
forces which act along the axes of a rectangular system of coordinates. 



GENERAL PROBLEMS. 

1. A particle is in equilibrium under the action of the forces P, Q, and 
R. Prove that 

P = Q = R 

sin (Q, R) sin (P, R) sin (P, Q) ' 

where (Q, R), etc., denote the angles between Q and R, etc. 

2. Two particles of weights W i and W 2 rest upon a smooth sphere of 
radius a. The particles are attached to the ends of a string of length I, 
which passes over a smooth peg vertically above the center of the sphere. 
If h is the distance between the peg and the center of the sphere, find (1) 
the position of equilibrium of the particles, (2) the tensile force in the 
string, and (3) the reaction of the sphere. 

3. The lengths of the mast and the boom of a derrick are a and b 
respectively. Supposing the hinges at the lower end of the boom and the 
pulley at the upper end to be smooth, find the angle the boom makes 
with the vertical when a weight W is suspended in equilibrium. 

4. Find the tensile force in the chain and the compression in the boom 
of the preceding problem. 

5. Two rings of weights W t and W 2 are held on a smooth circular 
wire in a vertical plane by means of a string subtending an angle 2 a at 
the center. Show that the inclination of the string to the horizon is 
given by 

tan "=Jv^ tana - 

6. A bridge, Fig. (a), of 60-foot span and 40-foot width has two queen- 
post trusses 9 feet deep. ' Each truss is divided into three equal parts by two 



28 



ANALYTICAL MECHANICS 



posts. What are the stresses in the different parts of the trusses when 
there is a load of 150 pounds per square foot of floor space? 



k 20- - ->j< 20- - -»u 20-- -*r 





Fig. (a). 



Fig. (b). 



7. Find the force in one of the members of the truss of figure (b). 

8. A weight rests upon a smooth inclined plane, supported by two equal 
strings the upper ends of which are fastened to two points of the plane in 
the same horizontal line. Find the tensile force in the strings and the 
reaction of the plane. 

9. In the preceding problem suppose the plane to be rough. 

10. A particle is suspended by a string which passes through a smooth 
ring fastened to the highest point of a circular wire in a vertical plane. 
The other end of the string is attached to a smooth bead which is movable 
on the wire. Find the position of equilibrium supposing the bead and 
the suspended body to have equal weights. 

11. A particle is in equilibrium on a rough inclined plane under the 
action of a force which acts along the plane. If the least magnitude of the 
force when the inclination of the plane is a equals the greatest magnitude 



when it is a 2 , show that <f> 



ai a * , where <f> is the angle of friction. 



12. Two weights W\ and W 2 rest upon a rough inclined plane, con- 
nected by a string which passes through a smooth pulley in the plane. 
Find the greatest inclination the plane can be given without disturbing 
the equilibrium. 

13. Two equal weights, which are connected by a string, rest upon a 
rough inclined plane. If the direction of the string is along the steepest 
slope of the plane and if the coefficients of friction are Mi and jjl 2 , find the 
greatest inclination the plane can be given without disturbing the equi- 
librium. 

14. In the preceding problem find the tensile force in the string. 

15. One end of a uniform rod rests upon a rough peg, while the other 
end is connected, by means of a string, to a point in the horizontal plane 
which contains the peg. When the rod is just on the point of motion it 



EQUILIBRIUM OF A PARTICLE 29 

is perpendicular to the string. Show that 2 I = fia, where I is the length 
of the string, a that of the rod, and /jl the coefficient of friction. 

16. A particle resting upon an inclined plane is at the point of motion 
under the action of a force F, which acts downward along the plane. If 
the angle of elevation of the plane is changed from a x to a 2 and the 
direction of the force reversed the particle will barely start to move up the 
plane. Express ju in terms of a.\ and a 2 . 

17. A string, which passes over the vertex of a rough double inclined 
plane, supports two weights. Show that the plane must be tilted through 
an angle equal to twice the angle of friction, in order to bring it from the 
position at which the particles will begin to move in one direction to the 
position at which they will begin to move in the opposite direction. 

18. Three equal spheres are placed on a smooth horizontal plane and 
are kept together by a string, which surrounds them in the plane of their 
centers. If a fourth equal sphere is placed on top of these, prove that the 

W 

tensile force in the string is — — , where W is the weight of each sphere. 

3 v6 

19. Three equal hemispheres rest with their bases upon a rough hori- 
zontal plane and are in contact with one another. What is the least value 
of fj. which will enable them to support a smooth sphere of the same radius 
and material ? 

20. If the center of gravity of a rod is at a distance a from one end and 
b from the other, find the least value of /jl which will allow it to rest in 
all positions upon a rough horizontal ground and against a rough vertical 
wall. 

21. A string, which is slung over two smooth pegs at the same level, 
supports two bodies of equal weight W at the ends, and a weight W at 
the middle' by means of a smooth ring through which it passes. Find 
the position of equilibrium of the middle weight. , 



CHAPTER III, 



EQUILIBRIUM OF RIGID BODIES. 

TRANSLATION AND ROTATION. 

32. Rigid Body. — There are problems in which bodies 
cannot be treated as single particles. In such cases they are 
considered to be made up of a great number of discrete par- 
ticles. A body is said to be rigid if the distances between 
its particles remain unchanged whatever the forces to which 
it may be subjected. There are no bodies which are strictly 
rigid. All bodies are deformed more or less under the action 
of forces. But in most problems discussed in this book ordi- 
nary solids may be treated as rigid bodies. 

33. Motion of a Rigid Body. — A rigid body may have two 
distinct types of motion. When the body moves so that its 
particles describe straight paths it 
is said to have a motion of trans- 
lation. Evidently the paths of the 
particles are parallel, Fig. 20. If 
the particles of the body describe 
circular paths it is said to have a 
motion of rotation. The planes of 
the circles are parallel, while their 
centers lie on a straight line per- 
pendicular to these planes, which 
is called the axis of rotation. The 
motion of a flywheel is a well-known example of motion of 
rotation. Suppose A, Fig. 21, to be a rigid body which is 
brought from the position A to the position A' by a motion 
of rotation about an axis through the point perpendicular 
to the plane of the paper, then the paths of its particles 

30 




Fig. 20. 



EQUILIBRIUM OF RIGID BODIES 



31 




Fig. 21. 



are arcs of circles whose planes 
are parallel to the plane of the 
paper and whose centers lie 
on the axis of rotation. 

34. Uniplanar Motion. — 
When a rigid body moves so 
that each of its particles re- 
mains at a constant distance 
from a fixed plane the motion 
is said to be uniplanar. The 
fixed plane is called the guide plane. 

35. Theorem I. — Uniplanar motion of a rigid body consists 
of a succession of infinitesimal rotational displacements. 

Suppose the rigid body A, Fig. 22, to describe a uniplanar 
motion parallel to the plane of the paper and let A and A' 
be any two positions occupied by the body. Then it may 
be brought from 4 to i' by a rotational displacement 
about an axis the position of Q 

which may be found in the fol- 
lowing manner. Let P and Q 
be the positions of any two 
particles of the body in a plane 
parallel to the plane of the 
paper when the body is at the 
position A, and P' and Q f be the 
positions of the same particles 
when the body occupies the po- 
sition A'. Then the desired axis is perpendicular to the 
plane of the paper and passes through the point of 
intersection of the perpendicular bisectors of the lines 
PQ and P'Q', drawn in the plane determined by these 
lines. 

Therefore the body can be brought from any position A 
to any other position A' by a single rotational displacement. 
The actual motion between A and A' will be, in general, 




Fig. 22. 



32 ANALYTICAL MECHANICS 

quite different from the simple rotation by which we accom- 
plished the passage of the body from one of these positions 
to the other. But the result, which we have just obtained, 
is true not only for positions which are separated by finite 
distances but also for positions which are infinitely near 
each other. Therefore by giving the body infinitesimal 
rotational displacements about properly chosen axes it may 
be made to assume all the positions which it occupies during 
its actual motion. 

36. Instantaneous Axis. — As the body is made to occupy 
the various positions of its actual motion the axis of rota- 
tion moves at right angles to itself and generates a cylinder 
whose elements are perpendicular to the guide plane. 
The elements of the cylinder are called instantaneous axes, 
because each acts as the axis of rotation at the instant when 
the body occupies a certain position. The curve of inter- 
section of the cylinder and the guide plane is called the 
centrode. 

The motion of a cylinder which rolls in a larger cylinder 
is a simple example of uniplanar motion. In this case the 
common element of contact is the instantaneous axis. As 
the cylinder rolls different elements of the fixed cylinder 
become the axis of rotation. 

Motion of translation and motion of rotation are special 
cases of uniplanar motion. In motion of translation the 
axis of rotation is infinitely far from the moving body. In 
rotation the cylinder formed by the instantaneous axes 
reduces to a single line, i.e., the axis of rotation. 

37. Theorem II. — Rotation about any axis is equivalent to a 
rotation through the same angle about a parallel axis and a 
translation in a direction perpendicular to it. 

The truth of this theorem will be seen from Fig. 23, where 
the rigid body A is brought from the position A to the posi- 
tion A' by a single rotation about an axis through the point 
perpendicular to the plane of the paper. This displace- 



EQUILIBRIUM OF RIGID BODIES 



33 



merit may be produced also by rotating the body to the 
position A" and then translating it to the position A' . 




Fig. 23. 



PROBLEMS. 

, 1. Show that in theorem II the order of the rotation and of the trans- 
lation may be changed. 

2. Show that the converse of theorem II is true. 

38. Theorem III. — The most general displacement of a rigid 
body can be obtained by a single translation and a single 
rotation. 

Let A and A' be any two positions occupied by the rigid 
body and P and P' be the corresponding positions of any one 




Fig. 24. 



of its particles. Then the body may be brought from A to 
A' by giving it a motion of translation which will bring the 
particle from P to P' and then rotating the body about a 
properly chosen axis through P'. A special case of this 



34 



ANALYTICAL MECHANICS 



theorem is illustrated in Fig. 24, where the direction of the 
translation is perpendicular to the axis of rotation. 

39. Theorem IV. — The most general displacement of a rigid 
body can be obtained by a displacement similar to that of a 
screw in its nut, that is, by a rotation about an axis and a 
translation along it. 

This theorem states that the axis of rotation of the last 
theorem can be so chosen that the translation is along the 
axis of rotation. Let PP' , Fig. 25, be the path of any point 
of the body described during 
the translation and BB be the 
line about which the body is 
rotated. Draw CC through 
P parallel to BB and drop the 
perpendicular P'P" upon CC. 
The displacement may be ac- 
complished now in the fol- 
lowing three stages. First: 
translate the body along the 
line CC until the point which 
was at P arrives at P" . 
Second: translate the body along P"P' until the point 
arrives at P' . Third: rotate the body about BB until it 
comes to the desired position. But by theorem II the 
last two operations can be accomplished by a single rotation 
about CC. Therefore the desired displacement can be 
obtained by a translation along and a rotation about the 
line CC. 

Evidently the last theorem holds for infinitesimal dis- 
placements as well as for finite displacements; therefore 
however complicated the motion of a rigid body it can be 
reproduced by a succession of infinitesimal screw-displace- 
ments, each displacement taking the body from one position 
which it has occupied during the motion to another position 
infinitely near it. Thus at every instant of its motion the 




EQUILIBRIUM OF RIGID BODIES 35 

rigid body is displaced like a screw in its nut. In general 
the pitch and the direction of the axis of the screw-motion 
change from instant to instant/ In the case of the motion 
of a screw in its nut these do not change. 

Translation and rotation are special cases of screw- 
motion. When the pitch of a screw is made smaller and 
smaller it advances less and less during each revolution. 
Therefore if the pitch is made to vanish the screw does 
not advance at all when it is rotated. Thus rotation is a 
special case of screw-motion in which the pitch is zero. 
On the other hand as the pitch of the screw is made greater 
and greater the screw advances more and more during each 
revolution. Therefore at the limit when the pitch is in- 
finitely great the motion of the screw becomes a motion of 
translation. Thus translation is a special case of screw- 
motion in which the pitch is infinitely great. 

LINEAR AND ANGULAR ACTION. TORQUE. 

40. Two Types of Action. — We have seen that a rigid body 
may have two different and independent types of motion, 
namely, motion of translation and motion of rotation. 
These motions are the results of two independent and 
entirely different kinds of actions to which a rigid body 
is capable of being subjected. We will differentiate between 
these two types of action by adding the adjectives " linear" 
and " angular" to the term "action." Thus the action which 
tends to produce translation will be called linear action and 
that which tends to produce rotation angular action. 

41. Torque. — The vector magnitude which represents the 
angular action of one body upon another is called torque. 

42. Couple. — Although a single force is not capable of pro- 
ducing the effect of a torque upon a rigid body, two or more 
external forces will produce it when properly applied. The 
simplest system of forces which is capable of producing 
rotation is known as a couple. It consists of two equal and 
opposite forces which are not in the same line, Fig. 26. 



36 



ANALYTICAL MECHANICS 



It is evident from Fig. 26 that a couple is capable of 
giving a rigid body a motion of rotation. But this is not 
enough to show that the effect produced by a couple is the 
same as that produced by a torque. We must show also 
that the couple is not capable of producing a motion of 
translation. Consider the rigid body A, Fig. 27, which is 
acted upon by a couple. Suppose the couple did tend to 





Fig. 26. 



Fig. 27. 



produce a translation in a direction BB'. Then pass through 
the body a smooth bar of rectangular cross-section in the 
direction of the supposed motion, so that the body is free 
to move along the bar but not free to rotate. When this 
constraint is imposed upon the rigid body it behaves like 
a particle and therefore cannot be given a motion by two 
equal and opposite forces. But since any motion in the 
direction BB ' is not affected by the presence of the bar, 
the assumption that the couple produces a motion of trans- 
lation along BB' must be wrong. Hence we see that when 
the bar is taken out the motion due to the couple will be 
one of pure rotation. 

43. Measure of Torque. — When a rigid body is in equi- 
librium under the action of two couples it is always found 
that the product of one of the forces of one couple by the 
distance apart of the forces of the same couple equals the 



EQUILIBRIUM OF RIGID BODIES 



37 



corresponding product for the other couple. In order, for 
instance, that the rigid body A, Fig. 28, be in equilibrium, 
we must have 

FD = F'D'. 
Therefore the product FD is 
the measure of the torque of 
the couple formed by the forces 
F and — F, the lines of action 
of which are separated by the 
distance D. Thus denoting 
the torque of a couple by G, 
we have 

G=FD. 

The distance D is called the arm of the couple and the plane 
of the forces the plane of the couple. 

44. Unit Torque. — The torque of a couple whose forces 
are one pound each and whose arm is one foot is the unit of 
torque. The symbol for the unit torque is the lb. ft. 

45. Vector Representation of Torque. — Torque is a vector 
magnitude and is represented by a vector which is perpen- 

F*f G 




(I) 




Fig. 29. 



dicular to the plane of the couple. The vector points away 
from the observer when the couple tends to rotate the body 
in the clockwise direction and points towards the observer 
when it tends to rotate the body in the counterclockwise 
direction, Fig. 29. In the first case the torque is considered 
to be negative and in the second case positive. 



38 



ANALYTICAL MECHANICS 



46. Equal Couples. — Two cou- 
ples are equal when the vectors 
which represent their torques are 
equal in magnitude and have the 
same direction. The three couples 
in Fig. 30 are equal if G x = G 2 = G 3 . 

Resultant of two couples is a 
third couple, whose torque is the 
vector sum of the torques of the 
given couples. 




Fig. 30. 



PROBLEMS. 

1. Find the direction and magnitude of the resultant torque of three 
equal couples the forces of which act along the edges of the bases of a 
right prism. The bases of the prism are equilateral triangles. 

2. In the preceding problem let the forces have a magnitude of 15 
pounds each, the length of the prism be 2 feet and the sides of the bases 
10 inches. 

3. In problem 1 suppose the prism to have hexagonal bases. 
L 4. In problem 2 suppose the prism to be hexagonal. 

5. A right circular cone, of weight W and angle 2 a, is placed in a 
circular hole of radius r, cut in a horizontal table. Assuming the coeffi- 
cient of friction between the cone and the table to be m, find the least 
torque necessary to rotate the former about its axis. . , 

47. Moment of a Force. — The most common method of 
giving a rigid body a motion of rotation is to put an axle 
through it and to apply to it a 
force which acts in a plane per- 
pendicular to the axle. The 
rotation is produced by the 
couple formed by the applied 
force and the reaction of the 
axle. The torque due to the 
couple equals the product of 
the applied force by the shortest distance from the axle 
to the line of action of the force. It is often more con- 




Fig. 31. 



EQUILIBRIUM OF RIGID BODIES 39 

venient to disregard the reaction of the axle. When this 
is done the torque of the couple is called the moment of the 
force applied. Therefore the moment of a force about an 
axis equals the product of the force by its lever-arm. The lever- 
arm of a force is the shortest distance between the axis and 
the line of action of the force. In Fig. 31 the moment of 
F about the axis through the point and perpendicular to 
the plane of the paper is 

G = Fd, (II) 

where d is the lever-arm. 

PROBLEMS. 

1. Prove that the moment of a force about an axis equals the moment 
of its component which lies in a plane perpendicular to the axis. 

2. Prove that the sum of the moments of the forces of a couple about 
any axis perpendicular to the plane of the couple is constant and equals 
the torque of the couple. 

48. Degrees of Freedom of a Rigid Body. — A rigid body 
may have a motion of translation along each of the axes of 
a rectangular system of coordinates and at the same time it 
can have a motion of rotation about each of these axes. 
Therefore a rigid body has six degrees of freedom, three of 
translation and three of rotation. When one point in it is 
constrained to move in a plane the number of degrees of 
freedom is reduced to five. When the point is constrained 
to move in a straight line the number becomes four. When 
the point is fixed the body has only the three degrees of 
freedom of rotation. If two points are fixed the body can 
only rotate about the line joining the two points. There- 
fore its freedom is reduced to one degree. When a third 
point, which is not in the line determined by the other two, 
is fixed the body cannot move at all, that is, it has no 
freedom of motion. 

49. The Law of Action and Reaction. — The law from which 
the conditions of equilibrium of a particle were obtained is a 



40 ANALYTICAL MECHANICS 

universal law applicable to all bodies under all conditions; 
therefore it is applicable to rigid bodies as well as to single 
particles. But since rigid bodies may be subject to two 
distinct types of action the law may be stated in the fol- 
lowing form. 

The sum of all the linear and angular actions to which 
a body or a part of body is subject at any instant vanishes. 

2(A, + AJ = (A') 

But since the two types of action are independent of each 
other the sum of each type must vanish when the combined 
sum vanishes. Therefore we can split the law into the fol- 
lowing two sections. 

To every linear action there is an equal and opposite 
linear reaction, or, the sum of all the linear actions to 
which a body or a part of body is subject at any instant 
vanishes: 

2A, = (A,) 

To every angular action there is an equal and opposite 
angular reaction, or, the sum of all the angular actions 
to which a body or a part of body is subject at any in- 
stant vanishes: 

2A a = (A a ) 

50. Conditions of Equilibrium of a Rigid Body. — If we replace 
the term " linear action" in the first section of the law by the 
word " force" and the term " angular action " in the second 
section of the law by the word "torque" we obtain the two 
conditions which must be satisfied in order that a rigid body 
be in equilibrium. Thus, in order that a rigid body be in 
equilibrium the following conditions must be satisfied. 

First. The sum of all the forces acting upon the rigid body 
must vanish, that is, if Fi, F 2 , • • • F n denote all the forces 
acting upon the body then the vector equation 

Fi+F 2 + ■ • • *F n *=6 (HI) 

must be satisfied. 



EQUILIBRIUM OF RIGID BODIES 41 

Second. The sum of all the torques acting upon the rigid 
body must vanish, that is, if Gi, G 2 , • ■ ■ O n denote all the 
torques acting upon the body then the vector equation 

Gi+G 2 + • • • +G„=0 (IV) 

must be satisfied. 

The following forms of the statement of these two condi- 
tions are better adapted for analysis. 

First. The algebraic sum of the components of all the forces 
along each of the axes of a rectangular system of coordinates 
must vanish, that is, 

2X=X 1 + X 2 + • • • +X n =0. 
2Y=Y 1 +Y 2 + • • • +F n =0. 
2Z ^ Zi + Z 2 + • • • + Z n = 0. 

Second. The algebraic sum of the components of all the 
torques about each of the axes of a system of rectangular coor- 
dinates must vanish, that is, 

2G X =G/ + (V , + ■ • • + <£° = 0. 



(V) 



v ^ y i ^y i ■ ^v 

2G 2 = GJ + 6," + • • • + Gi"> = 0. 



(VI') 



51. Coplanar Forces. — If two or more forces act in the same 
plane they are said to be coplanar. If a system of coplanar 
forces act in the xy-p\sme then the conditions of equilibrium 
reduce to the following equations: 

2X^X^X2+ • ■ • +X n =0,| () 

XY^Y,+ Y 2 + • • • +F n =0,j { } 

XG^F^ + F 2 d 2 + • • • + F n d n = 0, (VI) 

where d h a\, . . . , d n are the lever-arms of the forces Fi, 
F 2 , . . . F n , respectively, about any axis which is perpen- 
dicular to the plane of the forces. The ^-components of 
the forces and the x- and ^/-components of the moments 
vanish identically. Consequently they need not be con- 
sidered. 



42 ANALYTICAL MECHANICS 

52. Transmissibility of Force. — A force which acts upon a 
rigid body may be considered to be applied to any particle 
of the body which lies on the line of action of the force. In 
order to prove this statement consider the rigid body A, 
Fig. 32, which is in equilibrium under the action of the two 



Fig. 32. 

equal and opposite forces F and — F. Nov/ suppose we 
change the point of application of F, without changing 
either its direction or its line of application. Evidently 
the equilibrium is not disturbed, because by moving F in its 
line of action we neither changed the sum of the forces nor 
the sum of their moments about any axis. Therefore the 
line of action of a force is of importance and not its point 
of application. 

53. Internal Forces. — Internal forces do not affect the equi- 
librium of a rigid body. This is a direct consequence of the 
law of " action and reaction." Since by definition the in- 
ternal forces are due to the interaction between the particles 
of the system these forces exist in equal and opposite pairs, 
therefore mutually annul each other. 

ILLUSTRATIVE EXAMPLES. 

1. A uniform beam rests with its lower end on a smooth horizontal 
ground and its upper end against a smooth vertical wall. The beam is 
held from slipping by means of a string which connects the foot of the 
beam with the foot of the wall. Find the tensile force in the string and 
the reactions at the ends of the beam. 

There are four forces acting upon the beam, i.e., the two reactions, Ri 
and R 2 , the tensile force T and the weight W. Since both the ground and 
the wall are supposed to be smooth, Ri is normal to the ground, and R 2 



EQUILIBRIUM OF RIGID BODIES 



43 



to the wall. Therefore denoting the lengths of the beam and the string 
by I and a respectively, we have 

SZ m R 2 - T = 0, 

zZY sb & - W= 0, 

XG '^ -R 2 l sin a -\-W^cos a = 0, 

where zZG ' denotes the sum of the mo- 
ments of the forces about an axis through 
the point 0' perpendicular to the a^/-plane. 
Solving the last three equations we have 
Ri = W, 



and 



w 

R 2 = — cot a 

Z 

W a 




W \ | Rl 


2 Vl 2 -a 2 


*t\ 


T W a 





T C o' 


2 VI 2 -a 2 


Fig. 33. 



■X 



Discussion. — It should be noticed that in taking the moments the 
axis was chosen through the point 0' in order to eliminate the moments 
of as many forces as possible and thus to obtain a simple equation. 

The reaction Ri is independent of the angular position of the beam 
and equals the weight W. On the other hand R 2 and T vary with a. 



When a 



- both R 2 and T vanish. As a is diminished from - to 0, R 2 

Zi Zi 



and T increase indefinitely. 

2. A ladder rests on a rough horizontal ground and against a rough 
vertical wall. The coefficient of friction between the ladder and the 
ground is the same as that between the ladder and the wall. Find the 
smallest angle the ladder can make with the horizon without slipping. 

There are three forces acting on the ladder, i.e., its own weight W and 
the two reactions Ri and R 2 . Replacing Ri and R 2 by their components 
and writing the equations of equilibrium we obtain 
zZX = F x - N 2 = 0, 
XY ^N 1 -\-F 2 -W = } 



2GV = F 2 l cos a + N 2 l sin a 



PT-cos a = 0, 

Zi 



where a is the required angle. 
We have further 



M = 



Ni 



N 2 



44 



ANALYTICAL MECHANICS 



Solving these we get 



Ni = 



1+M 5 



W. 



Ri = 



F,= 



Vl+M 2 

..5 



w, 



1+M : 



iV 2 = — - — W 
2 1 + m 2 ' 



#2 = 

tan a = 



Vl + m 2 ' 
1-M 2 
2m 




Fig. 34. 



Discussion. — The last expression gives the value of a for a given 
value of fx. When /* = 1, a = 0, therefore in this case the ladder will be 

in equilibrium at any angle between and - with the ground. Evidently 

this is true for any value of /jl greater than unity. 

3. Find the smallest force which, when applied at the center of a 
carriage wheel, will drag it over an obstacle. 

The forces acting on the wheel are, its weight W, the required force F, 
and the reaction R. Since the first two meet at the center of the wheel, 
the direction of R must pass through the center also. Take the coordinate 
axes along and at right angles to R, as shown in Fig. 31, and let F make 
an angle 6 with the a;-axis. Then the equations of equilibrium become 

XX ^Fcosd - R + Wcosa = 0, 
SF =Fsind - IT sin a = 0, 
XG '= W -asm a- Fsin6-a = 0. 

From either of the last two equations we get 

F== sma w 

sin v 

Since W and a are fixed F can be changed only by changing 0. Therefore 
the minimum value of F is given by the maximum value of sin 0, i e., 



6 = -, which makes 



F = W sin a. 



EQUILIBRIUM OF RIGID BODIES 45 

a — h 



From the figure we obtain cos a = 

Kb 

therefore sin a = - Vh (2 a — h), 

a 

and F = — Vh(2a-h). 




Fig. 35. 

Since cos 6 = the first equation of equilibrium gives 

R = W cos a 



Discussion. — It will be observed that the first two of the equations 
of equilibrium are sufficient to solve the problem. 

When h is zero, F = and R = W. On the other hand when h = a, 
F = W and R = 0. 

PROBLEMS. 

1. Prove that the true weight of a body is the geometric mean between 
the apparent weights obtained by weighing it in both pans of a false 
balance. 

2. A uniform bar weighing 10 pounds is supported at the ends. A 
weight of 25 pounds is suspended from a point 20 cm. from one end. 
Find the pressure at the supports if the length of the bar is 50 cm. 

3. A uniform rod which rests on a rough horizontal floor against a 
smooth vertical wall is on the point of slipping. Find the reactions at 
the two ends of the rod. 



46 ANALYTICAL MECHANICS 

4. A body is suspended from the middle of a uniform rod which passes 
over two fixed supports 6 feet apart. In moving the body 6 inches nearer 
to one of the supports the pressure on the support increases by 100 
pounds. What is the weight of the body if 5 pounds is the weight of 
the rod? 

5. A uniform rod of length a and weight W is suspended by two strings 
having lengths k and Z 2 . The lower ends of the strings are attached to the 
ends of the rod, while the upper ends are tied to a peg. Find the tensile 
force in the strings. 

6. A safety valve consists of a cylinder with a plunger attached to a 
uniform bar hinged at one end. The plunger has a diameter of \ inch 
and is attached to the bar at a distance of 1 inch from the hinge. The 
bar is 2 feet long and weighs 1 pound. How far from the hinge must a 
slide-weight of 2 pounds be set if the steam is to blow off at 120 pounds 
per square inch? 

7. The two legs of a stepladder are hinged at the top and connected 
at the middle by a string of negligible mass. Find the tensile force in the 
string and the pressure on the hinges when the ladder stands on a smooth 
plane. The weight of the ladder is W, the length of its legs I, and the 
length of the string a. 

8. A uniform rod rests on two inclined planes making angles of ai and 
cli with the horizon. Find the angle which the rod makes with the 
horizon and the pressure on the planes. 

9. A rectangular block is placed on a rough inclined plane whose in- 
clination is gradually increased. If the block begins to slide and to turn 
about its lowest edge simultaneously find the coefficient of friction. 

10. A uniform rod rests with one end against a rough vertical wall 
and the other end connected to a point in the wall by a string of equal 
length. Show that the smallest angle which the string can make with 

'-©■ 

11. A uniform rod is suspended by a string which is attached to the 
ends and is slung over a smooth peg. Show that in equilibrium the rod 
is either horizontal or vertical. 

12. A ladder 25 feet long and weighing 50 pounds rests against a 
vertical wall making 30° with it. How high can a man weighing 150 
pounds climb up the ladder before it begins to slip? The coefficient of 
friction is 0.5 at both ends of the ladder. 

13. A rod of negligible weight rests wholly inside a smooth hemispheri- 
cal bowl of radius r. A weight W is clamped on to the rod at a point 
whose distances from the ends are a and b. Show that the equilibrium 



the wall is tan' 



EQUILIBRIUM OF RIGID BODIES 



47 



position of the rod is given by sin = 



a — b 



where 6 is the angle it 



2Vr 2 -ab' 
makes with the plane of the brim of the bowl which is horizontal. 

14. Prove that when a rigid body is in equilibrium under the action of 
three forces their lines of action lie in the same plane and intersect at the 
same point. 

15. Find the forces which tend to compress or extend the different 
members of the following cranes. 



1,750 lbs 




16. Supposing the weights of the following figures to be in equilibrium 
find their relative magnitudes. The circles which are tangent to other 
circles represent gears. 




54. Resultant of a System of Forces Acting upon a Rigid Body. 
— We have already shown that the most general displacement 
of a rigid body consists of a translation along, and a rotation 
about, a certain line. Therefore such a displacement can be 
prevented by a single force opposed to the translation and 
a single torque opposed to the rotation. Thus a single force 
and a single torque can be found which will keep a rigid body 
in equilibrium against the action of any system of forces. 



48 ANALYTICAL MECHANICS 

The resultant of a system of forces consists, therefore, of a 
single force and a single torque which, when reversed, will 
keep the rigid body in equilibrium against the action of the 
given system of forces. 

55. Resultant of Coplanar Forces Acting upon a Rigid Body. — 
Let Pi, F 2 , . . . F n denote the given forces and let the xy- 
plane be their plane of action. Then, if R, X, and Y denote 
the resultant force and its components, respectively, we have 



Z = Xi + X 2 + ■ • 


• + x„ 


Y=Y 1 +Y 2 + • • 


• +Y n , 


R = Vx 2 +Y 2 , 




Y 

tan = —> 
A 





(VII) 

. • (VIII) 

and tan B = ~ > (IX) 

A 

where the terms in the right-hand members of the first two 
equations are the components of the given forces, and 9 is 
the angle R makes with the x-axis. 

On the other hand if G denotes the resultant torque and 
d h ch, . . . , d n denote the distances of the origin from the 
lines of action of the forces, then 

G = F 1 d 1 + F 2 d 2 + • • • + F n d n . (X) 

If we represent this torque by the moment of the resultant 
force about the 2-axis, then 

RD=F 1 d 1 + F 2 d 2 + • • • +F n d n 
2Fd 



or D = 



(XI) 



R 

gives the distance of the line of action of the resultant force 
from the origin. 

ILLUSTRATIVE EXAMPLE. 

Find the resultant of the six forces acting along the sides of the hexa- 
gon of Fig. 36. 

Taking the sum of the components along the x and y directions, we 
have 



EQUILIBRIUM OF RIGID BODIES 



49 



X = 2F + 3Fcosf-i2Fcosf-F-2^cosf + ^cosf 

o o o o 

= F. 

F = 0-3^sinJ-2^sinf + + 2Fsinf + ^sinf 

o o o o 

= -fVs. 




... R = V^ 2 + 3 F 2 
= 2F 
and tan d = — V3. 

Therefore the resultant force has 
a magnitude 2 F and makes an angle 
of —60° with the x-axis. 

Taking the moments about an 
axis through the center of the hexa- 
gon, we obtain Fig. 36. 
RD = (2F + 3F + 2F + F + 2F + F)a 
= 11 Fa, 
therefore D = 5.5a, 

where a is the distance of the center from the lines of action of the forces. 

56. Resultant of a System of Parallel Forces. — Let R be the 

resultant of the parallel forces Fi, F 2 , . . ., F„, which act 
upon a rigid body. Then, since the forces are parallel, the 
resultant force equals the algebraic sum of the given forces. 

Thus 

R = Fl +F 2+ . . . + Fni 

and RD = F 1 di + F 2 d 2 + • • • + F n d n . 

Now take the 2-axis parallel to the forces and let x { and y { 
denote the distances of F { from the yz-plane and the xz- 
plane, respectively. Then the last equation may be split 
into two parts, one of which gives the moments about the 
x-axis and the other about the y-axis. Thus, 

Rx^=F 1 Xi + F 2 x 2 -\- • • • +F n x n , ) 
Ry = F l y 1 + F 2 y 2 + • • • +F n y H ,) 

where x and y are the coordinates of the point in the xy- 
plane through which the resultant force passes. In other 



50 



ANALYTICAL MECHANICS 



words, (x,y) is the point of application of the resultant force. 
The resultant force is evidently parallel to the given forces. 
The last two equations may be written in the following 
forms 

ZFx 



X R 
y R 



(XIII) 



ILLUSTRATIVE EXAMPLE. 

Find the resultant of two parallel forces which act upon a rigid body 
in the same direction. 

Let the ?/-axis be parallel to the 
forces. 
Then R = F X + F 2 , 



and 



or 



x = 



F, 
F 2 



F&x + F 2 x 2 
F, + F 2 ' 
x 2 — x 




X — X\ 

But since x 2 — x and x — x x are the 
distances of F2 and Fi from R, we have 

F\ = d_ 2 

F 2 dx 

or F x dx = F 2 d 2 . 

Therefore the distances of the resultant from the given forces are in- 
versely proportional to the magnitudes of the latter. 

PROBLEMS. 

1. Find the resultant force and the resultant torque due to the forces 
P, 2 P, 4 P and 2 P which act along the sides of a square, taken in order. 

2. Three forces are represented in magnitude and line of action by 
the sides of an equilateral triangle. Find the resultant force, taking the 
directions of one of the forces opposite to that of the other two. 

3. The lines of action of three forces form a right isosceles triangle of 
sides a, a, and a V2. The magnitudes of the forces are proportional to 
the sides of the triangle. Find the resultant force. 

4. The sum of the moments of a system of coplanar forces about any 
three points, which are not in the same straight line, are the same. Show 
that the system is equivalent to a couple. 



EQUILIBRIUM OF RIGID BODIES 51 

5. Three forces are represented in magnitude, direction, and line of 
action by the sides of a triangle taken in order ; prove that their resultant 
is a couple the torque of which equals, numerically, twice the area of the 
triangle. 

6. Three forces act along the sides of an equilateral triangle, find the 
condition which will make their resultant pass through the center of the 
triangle. 

FRICTION ON JOURNALS AND PIVOTS. 

57. Friction on Journal Bearing. — If the horizontal shaft 
of Fig. 38 fits perfectly in its bearing the friction which comes 
into play is a sliding friction, therefore the laws of sliding 
friction may be assumed to hold good. The most important 
of these laws is : the frictional force which comes into play 
is proportional to the normal reaction, that is, in the relation 

n is independent of N. We will assume therefore that this 
law holds at each point of the surface of contact and thus 
reduce the problem under discussion to one of sliding fric- 
tion. There is an important difference, however, between 
the problem of friction on journal bearing and the problems 
on friction which we have already discussed. In the present 
problem the normal reaction is not the same at the points 
of the surfaces in contact. We must apply, therefore, the 
laws of friction to small elements of surfaces of contact over 
which the normal reaction may be considered to be constant. 
Let the element of surface be a strip, along the length of 
the shaft, which subtends an angle dd at the axis of the shaft. 
Further let <iN be the normal reaction over this element of 
surface, and dF be the corresponding frictional force; then 
we have 

dF = m dN 

= /xp -l -add, 

where p is the normal reaction per unit area or the pressure, 
a is the radius of the shaft and I the length of the bearing. 



52 



ANALYTICAL MECHANICS 



Therefore the total frictional force and the total frictional 
torque are, respectively, 



and 



F = fxal f pdd 

T 

G = fxaH f 2 p do. 



In order to carry out the integral of the foregoing expressions 
we have to make some assumption with regard to the nature 




Fig. 38. 

of dependence of .p upon 6. But whatever the relation 
between p and it is ^obvious that the sum, over all the sur- 
face of contact, of the vertical component of the normal 
reaction must equal to the load which rests upon the bear- 
ing. If P denotes this load, then p must satisfy ^ the condi- 
tion 



£ 



p sin • dA 



= at j psin.6 dd, 
where A is the total area of contact. 

ILLUSTRATIVE EXAMPLE. 

The normal pressure on the bearing is given by the relation p 
find the total frictional force and the total frictional torque. 



Posin 0: 



EQUILIBRIUM OF RIGID BODIES 53 

Substituting the given value of p in the expression for F we obtain 

IT 

sin. 6 dQ 

r "" ? = fxalpo. 

In order to determine po in terms of the total load on the bearings we 
make p satisfy the condition 



P = alCp sin • dB. 



Substituting the given value of p in the right-hand member of the pre- 
ceding equation we have 



P = alpof 2 sm 2 ddd 



•2 

'0 
_ TTdlpo 



4P 
Therefore F = ^P 

7T 

and G = — aP. 

x 

It will be observed that the total frictional force varies with the load and 
is independent of the radius and of the length of the bearing; in other 
words it is independent of the area of contact. 

PROBLEMS. 

1. Supposing the normal pressure to be the same at every point of the 
surfaces of contact, derive the expressions for the total frictional force and 
the resisting torque due to friction. 

2. Supposing the vertical component of the normal pressure at every 
point of the surfaces of contact to be constant, derive the expressions for 
the total frictional force and the resisting torque due to friction. 

3. Derive expressions for the total frictional force and the resisting 
torque upon the assumption that the normal pressure is given by the 
relation p = p sin 2 0. 

58. Friction on Pivots. — The problem of friction on 
pivots also is a problem of sliding friction. The feature 



54 



ANALYTICAL MECHANICS 




which distinguishes the pivot from the journal bearing is 

this : in the former the lever arm of the frictional force varies 

from point to point, while 

in the latter it is constant 

and equals the radius of the 

shaft. 

Let dN be the normal 
reaction upon dA, an ele- 
ment of area at the base of 
the flat-end pivot of Fig. 
39; then if dF denotes 
the corresponding frictional 
force, we have 

dF=tidN 
= fip • dA, 

where p is the normal pres- 
sure. Evidently p is con- 
stant; therefore we can write 



» p £ 



dA 




= -Ka l [ip. 



Fig. 39. 



The expression for the resisting torque due to the friction 
is obtained as follows : 



G 



-£ 

Jo Jo 



r-dF 



fxpdA 



r»p 



= it tip j r 2 dr 
= lira? up 
= I «mP, 
where P is the total load on the pivot. 



-dr 



EQUILIBRIUM OF RIGID BODIES 



55 



PROBLEMS. 

1. Derive an expression for the resisting torque due to friction in the 
collar-bearing pivot of the adjoining figure. 




i^^~^ 




2. Supposing the normal pressure to be constant, derive an expression 
for the resisting torque due to friction in the conical pivot of the adjoining 
figure. 

3. In the preceding problem suppose the vertical component of the 
normal pressure to be constant. 




4. In problem 2 suppose the horizontal component of the normal 
pressure to be constant. 

5. Taking the normal pressure to be constant derive an expression 
for the resisting torque, due to friction in the spherical pivot of the adjoin- 
ing figure. 



56 



ANALYTICAL MECHANICS 



6. Prove that the resisting torque due to friction is greater for a hollow 
pivot than for a solid pivot, provided that the load and the load per unit 
area are the same in both cases. 

7. Show that the resisting torque due to friction for a hemispherical 
pivot is about 2.35 times as large as that for a flat end pivot. 



ROLLING] FRICTION. 

59. Coefficient of Rolling Friction. — Consider a cylinder, 
Fig. 38, which is in equilibrium on a rough horizontal plane 
under the action of a force S- y 

In addition to this force the 
cylinder is acted upon by its 
weight and by the reaction of 
the plane. Applying the con- 
ditions of equilibrium we ob- 
tain 



2X = S-F=0, 

27 = -W+N=0, 
2G = ND-Sd=0, 

where F and N are the com- 
ponents of R, the reaction of 
the plane, while D and d are, 
respectively, the distances of 
the point of application of R and S from the point 0, about 
which the moments are taken. These equations give us 




Fig. 40. 



and 



r=Vf 2 +n 2 
= Vs 2 +w 2 , 
s 



D 



w 



d. 



(1) 

(2) 



If the cylinder is just on the point of motion then 

F=nN, 

a 

and consequently A t= w* ® 



EQUILIBRIUM OF RIGID BODIES 57 

Combining (2) and (3), we obtain 

D = fid. (XIV) 

The distance D is called the coefficient of rolling friction. 
Equation (XIV) states, therefore, that the coefficient of the 
rolling friction equals the coefficient of the sliding friction 
times the distance of the point of contact from the line of 
action of the force which urges the body to roll. 

60. Friction Couple. — It is evident from the above equa- 
tions that a change in the value of d does not affect the values 
of N and F, consequently it does not change the value of /x. 
This is as it should be, since, according to the laws of sliding 
friction, y depends only upon the nature of the surfaces in 
contact. A change in d, however, changes the value of D; 
in other words, it changes the point of application of R. 
When d = 0, that is, when S is applied at the point of con- 
tact, D= 0, in which case the body is urged to slide only. 
But when d is not zero the force S not only urges the body 
to slide but also to roll; therefore, in addition to the resist- 
ing force F, a resisting torque comes into play. This torque, 
which is due to the couple formed by N and W, is called 
friction couple. 

PROBLEMS. 

1. A gig is so constructed that when the shafts are horizontal the 
center of gravity of the gig is over the axle of the wheels. The gig rests 
on a perfectly rough horizontal ground. Find the least force which, 
acting at the ends of the shafts, will just move the gig. 

2. Find the smallest force which, acting tangentially at the rim of a 
flywheel, will rotate it. The weight and the radius of the flywheel, the 
radius of the shaft, and the coefficient of friction between the shaft and 
its bearings are supposed to be known. 

3. A flywheel of 500 pounds weight is brought to the point of rotation 
by a weight of 10 pounds suspended by a string wound around its rim. 
Find the coefficient of friction between the axle and its bearings. The 
diameters of the wheel and the axle are 10 feet and 8 inches, respectively. 



58 ANALYTICAL MECHANICS 

4. A wheel of radius a and weight W stands on a rough horizontal 
ground. If ^ is the coefficient of friction between the wheel- and the 
ground find the smallest weight which must be suspended at one end 
of the horizontal diameter in order to move the wheel. 



GENERAL PROBLEMS. 

1. A table of negligible weight has three legs, the feet forming an 
equilateral triangle. Find the proportion of the weight carried by the 
legs when a particle is placed on the table. 

2. A rectangular board is supported in a vertical position by two 
smooth pegs in a vertical wall. Show that if one of the diagonals is 
parallel to the line joining the pegs the other diagonal is vertical. 

3. A uniform rod rests with its two ends on smooth inclined planes 
making angles a and (3 with the horizon. Where must a weight equal to 
that of the rod be clamped in order that the rod may rest horizontally? 

4. A uniform ladder rests against a rough vertical wall. Show that 
the least angle it can make with the horizontal floor on which it rests is 

given by tan 6 == — , where /jl and yJ are the coefficients of friction 

2 ix 

for the floor and the wall respectively. 

5. A uniform rod is suspended by two equal strings attached to the 
ends. In position of equilibrium the strings are parallel and the bar is 
horizontal. Find the torque which will turn the bar, about a vertical 
axis, through an angle 6 and keep it in equilibrium at that position. 

6. The line of hinges of a door makes an angle a with the vertical. 
Find the resultant torque when the door makes an angle /3 with its equi- 
librium position. 

7. The lines of action of four forces form a quadrilateral. If the 
magnitude of the forces are a, b, c, d times the sides of the quadrilateral 
find the conditions of equilibrium. 

8. A force acts at the middle point of each side of a plane polygon. 
Each force is proportional to the length of the side it acts upon and is 
perpendicular to it. Prove that the polygon will be in equilibrium if all 
the forces are directed towards the inside of the polygon. 

9. A force acts at each vertex of a plane convex polygon in a direc- 
tion parallel to one of the sides forming the vertex. Show that if the 
forces are proportional to the sides to which they are parallel and if their 
directions are in a cyclic order their resultant is a couple. 

10. A uniform chain of length I hangs over a rough horizontal cylinder 
of radius a. Find the length of the portions which hang vertically when 



EQUILIBRIUM OF RIGID BODIES 59 

the chain is on the point of motion under its own weight, (1) when a is 
negligible compared with I, (2) when it is not negligible compared with I. 

11. Two equal weights are attached to the extremities of a string 
which hangs over a rough horizontal cylinder. Find the least amount 
by which either weight must be increased in order to start the system to 
move. The weight of the string is negligible. 

12. Three cylindrical pegs of equal radius and roughness are placed 
at the vertices of a vertical equilateral triangle the two lower corners of 
which are in the same horizontal line. A string of negligible weight is 
attached to two weights and slung over the pegs. Find the ratio of the 
weights if they are on the point of motion. 

13. A sphere laid upon a rough inclined plane of inclination a. is on the 
point of sliding. Show that the coefficient of friction is f tan a. 

14. A uniform ring of weight W hangs on a rough peg. A bead of 
weight w is fixed on the ring. Show that if the coefficient of friction 

W 

between the ring and the peg is greater than — the ring will 

VW* + 2wW 
be in equilibrium whatever the position of the bead with respect to the 
peg. 

15. A uniform rod is in equilibrium with its extremities on the interior 
of a rough vertical hoop. Find the limiting position of the rod. 

16. A weight W is suspended from the middle of a cord whose ends are 
attached to two rings on a horizontal pole. If w be the weight of each 
of the rings, /jl the coefficient of friction, and I the length of the cord, find 
the greatest distance apart between the rings compatible with equilibrium. 



CHAPTER IV. 
EQUILIBRIUM OF FLEXIBLE CORDS. 

61. Simplification of Problems. — The simplest phenome- 
non in nature is the result of innumerable actions and 
reactions. The consideration of all the factors which con- 
tribute to any natural phenomenon would require unlimited 
analytical power. Fortunately the factors which enter into 
dynamical problems are not of equal importance. Often the 
influence of one or two predominate, so that the rest can be 
neglected without an appreciable departure from the actual 
problem. Any one who attempts to solve a physical problem 
must recognize this fact and use it to advantage by repre- 
senting the actual problem by an ideal one which has only 
the important characteristics of the former. This was done 
in the last two chapters in which bodies were treated as single 
particles and rigid bodies, and thereby simplified the prob- 
lems without changing their character. 

The same procedure will be followed in discussing the 
equilibrium of flexible cords, such as belts, chains, and ropes. 
These bodies will be represented by an ideal cord of negli- 
gible cross-section and of perfect flexibility. The solution of 
the idealized problems gives us a close enough approxima- 
tion for practical purposes. If, however, closer approxima- 
tion is desired smaller factors, such as the effects of thickness 
and imperfect flexibility, may be taken into account. 

62. Flexibility. — A cord is said to be perfectly flexible if it 
offers no resistance to bending; in other words, in a perfectly 
flexible cord there are no internal forces which act in a 
direction perpendicular to its length. 

60 



EQUILIBRIUM OF FLEXIBLE CORDS 



61 



63. Suspension Bridge Problem. — The following are the 
important features of a suspension bridge which should be 
considered in order to simplify the problem: 

1. The weights of the cables and of the chains are small 
compared with that of the road-bed. { 

2. The road-bed is practically horizontal. 

3. The distribution of weight in the road-bed may be 
considered to be uniform. 

We can, therefore, obtain a sufficiently close approxima- 
tion if we consider an ideal bridge in which the cable and 
the chains have no weight and the distribution of weight in 
the road-bed is uniform in the horizontal direction. With 
these simplifications consider the forces acting upon that 
part of the cable which is between the lowest point and any 
point P, Fig. 41. 




Fig. 41. 



The forces are: The tensile force, T , acting horizontally 
at 0. The tensile force, T, acting along the tangent to the 
curve at P. The weight of that part of the bridge which 
is between and P. If w be the weight per unit length 
of the road-bed and x denotes the length OP, then the third 
force becomes wx. 

Therefore the conditions of equilibrium give 

SZ= -To + T 7 cos 0=0; /. Tcos6=T . (1) 

27 = -wx+ Tsin 0=0; /. Tsin = wx. (2) 



62 ANALYTICAL MECHANICS 

It is evident from equation (1) that the horizontal compo- 
nent of the tensile force is constant and equals T . Squaring 
equations (1) and (2) and adding we get 

T 2 = T 2 + w 2 x 2 . (3) 

Thus we see that the smallest value of T corresponds to 
x = and equals T , while its greatest value corresponds 
to the greatest value of x. If D denotes the span of the 
bridge then the greatest value of T, or the tensile force of 
the cable at the piers, is 



= \Jt< 



2 , ™ 2 D 2 
o 2 - 1 



4 

In order to find the equation of the curve which the cable 
assumes we eliminate T between equations (1) and (2). 
This gives 

tan d = — x. (4) 

To 

Substituting -y- for tan and integrating we get 
ax 

1 w 9 . 
V = - — ar+ c, 
y 2 T ' 

where c is the constant of integration. 

But with the axes we have chosen, y=0 when #=0, 
therefore c = 0. Thus the equation of the curve is 

u 2 To 
which is the equation of a parabola. 

Dip of the Cable. — Let H be the height of the piers 

above the lowest point of the cable. Then for x = — , y = H, 

therefore 

H=~D'. (6) 

It is evident from the last equation that the greater the 
tension the less is the sag. 



EQUILIBRIUM OF FLEXIBLE CORDS 



63 



Problem. A bridge is supported by two suspension cables. The 
bridge has a weight of 1.5 tons per horizontal foot and has a span of 400 
feet. Supposing the dip of the bridge to be 50 feet find the values of 
the tensile force at the lowest and highest points of the cable. 

64. Equilibrium of a Uniform Flexible Cord which is Sus- 
pended from Its Ends. — The problem is to determine the 
nature of the curve which a 
perfectly uniform and flexi- 
ble cable will assume when 
suspended from two points. 
Let AOB, Fig. 42, be the curve. 
Consider the equilibrium of 
that part of the cable which 
is between the lowest point 
and any other point P. 
The part of the cable which 
is under consideration is 
acted upon by the following three forces: 

The tensile force at the point 0, T . 

The tensile force at the point P, T. 

The weight of the cable between the points and P. 
Since the cable is perfectly flexible T and T are tangent to 
the curve. Therefore we have 

2X= -T + TcosO = 0, or T cos = T , (1) 

2F = -ws+Tsin d= 0, or T sin d = ws, (2) 

where w is the weight per unit length of the cable and s is 
the length of OP. 

Squaring equations (1) and (2) and adding we obtain 




'2 _ 



JV 



-\-w 2 s' 



Eliminating T between equations (1) and (2) we get 

To 

IV 



s = 



tan0, 



(3) 



(4) 



which is the intrinsic equation of the curve. 



64 



ANALYTICAL MECHANICS 



In order to express equation (4) in terms of rectangular 

coordinates we replace tan by -^ and obtain 

dx 



s = 



w dx 



(5) 



But ds 2 = dx 2 + dy 2 , therefore eliminating da; 'between this 
equation and equation (5) and separating the variables 



and then integrating 



, sds 

dy = —=^=> 

Vs 2 + a 2 



y = Vs 2 + a 2 + c, 



(6) 



where a = — and c is the constant of integration. 

Let the o>axis be so chosen that when s= 0, y = a, then 
c= 0. Therefore 



y = Vs 2 -\-a 2 , or s= ^y 2 — a'' 



(7) 



Differentiating equation (7), squaring and replacing ds 2 by 
(dx 2 + dy 2 ) we have , 



dx 2 +dy 2 = ^\. 
y 2 — a 2 



Solving for dx, 



dx= — 



v^/ 2 — a 2 
ady 



dy 



v-i V?- 
ady 



iVa 2 — y 2 



(8) 



where i = V — 1. Integrating equation (8) we get 



— = cos 1 - +c . 
a a 



EQUILIBRIUM OF FLEXIBLE CORDS 65 

But y — a, when x= 0, therefore d = 0. Thus we get 



%x 
y = a cos— ■> 


(9) 


= a cosh -, 
a 


(10) 


= l(ehe^\ 


(11) 


rp 1 wx wx\ 


(12) 



which are different forms of the equation of a catenary. 

Discussion. — Expanding equation (12) by Maclaurin's 
Theoremf we obtain 

In the neighborhood of the lowest point of the cable the 
value of x is small, therefore in (13) we can neglect all the 
terms which contain powers of x higher than the second. 
Thus the equation 

V=a+f a (14) 

represents, approximately, the curve in the neighborhood 
of the lowest point. It will be observed that (14) is the 
equation of a parabola. This result would be expected 
since the curve is practically straight in the neighborhood 
of O and consequently the horizontal distribution of mass 
is very nearly constant, which is the important feature of 
the Suspension Bridge problem. 

The nature of those parts of the curve which are removed 
from the lowest point may be studied by supposing x to be 

X 

large. Then since e a becomes negligible equation (11) re- 
duces to 

V=\e% ( 15 ) 

* See Appendix A. f See Appendix. 



66 



ANALYTICAL MECHANICS 



The curve, Fig. 43, denned 
by equation (15) is called an 
exponential curve. It has an 
interesting property, namely, 
its ordinate is doubled every 
time a constant value P is 
added to its abscissa. This 
constant is called the half-value 
period of the curve. The value 
of P may be determined in 
the following manner. By the 
definition of P and from equa- 
tion (15) we have 

2 y = -e 

Dividing (16) by (15) we get 




X + P 



(16) 



2 = e a , 
or P = a log e 2. 

Length of Cable. — In order to find the length in terms 
of the span eliminate y between equations (7) and (11). 
This gives 

~f * _*\ 

e a -e a (17) 



* + 2.3 a 2 + 2.3.4 



5 a' 



+ 



(18) 



where the right member of equation (18) is obtained by 
expanding the right-hand member of equation (17) by 
Maclaurin's Theorem. 

If D and L denote the span and the length of the cable, 
respectively, we have s = ^L when x=\D. Therefore sub- 
stituting these values of s and x in (18) and replacing a by 
its value we obtain 

^ 2 (* D+ f 8 f> + ---> (19) 



EQUILIBRIUM OF FLEXIBLE CORDS 67 

When the cable is stretched tight T is large compared with 
w. Therefore the higher terms of the series may be ne- 
glected and equation (1) be put in the following approxi- 
mate form. 



4+ilH- (20) 

l_ w^_ 
24 TV 



1 w 2 
Hence the increase in length due to sagging is — — 2 D 3 , 



approximately. 

PROBLEMS. 

1. A perfectly flexible cord hangs over two smooth pegs, with its ends 
hanging freely, while its central part hangs in a catenary. If the two 
pegs are on the same level and at a distance D apart, show that the total 
length of the string must not be less than De, in order that equilibrium 
shall be possible, where e is the natural logarithmic base. 

, 2. In the preceding problem show that the ends of the cord will be 
on the x-axis. 

3. Supposing that a telegraph wire cannot sustain more than the 
weight of one mile of its own length, find the least and the greatest sag 
allowable in a line where there are 20 poles to the mile. 

4. Find the actual length of the wire per mile of the line in the pre- 
ceding problem. 

5. The width of a river is measured by stretching a tape over it. 
The middle point of the tape touches the surface of the water while the 
ends are at a height H from the surface. If the tape reads S, show that 

/S 2 — H 2 
the width of the river is approximately i/ 

6. Show that the cost of wire and posts of a telegraph line is mini- 
mum if the cost of the posts is twice that of the additional length of wire 
required by sagging. The posts are supposed to be evenly spaced and 
large in number. 

7. A uniform cable which weighs 100 tons is suspended between two 
points, 500 feet apart, in the same horizontal line. The lowest point of 
the cable is 40 feet below the points of support. Find the smallest and 
the greatest values of the tensile force. 

8. In the preceding problem find the length of the cable. 

65. Friction Belts. — The flexible cord AB, Fig. 44, is in 
equilibrium under the action of three forces, namely, T 



68 



ANALYTICAL MECHANICS 



and T, which are applied at the ends of the cord, and the 
reaction of the rough surface of C, with which it is in con- 
tact. It is desired to find the relation between T and T 
when the cord is just on the point of motion towards T . 




Fig. 44. 



Consider the equilibrium of an element of that part of the 
cord which is in contact with the surface. The element 
is acted upon by the following three forces: 

The tensile force in the cord to the right of the element. 

The tensile force in the cord to the left of the element. 

The reaction of the surface. 

Let the tensile force to the left of the element be denoted 
by T, then the tensile force to the right may be denoted by 
T+ dT. On the other hand if R denotes the reaction of 
the surface per unit length of the cord, the reaction on the 
element is R ds, where ds is the length of the element. We 
will, as usual, replace R by its frictional component F and 
its normal component N. 



EQUILIBRIUM OF FLEXIBLE CORDS 69 

Taking the axes along the tangent and the normal through 
the middle point of the element and applying the conditions 
of equilibrium we obtain 

2X = (T + dT) cos— - T cos ^ - F ( - ds) = 0, 

2 Y = N ds - T sin ^ - (T + dT) sin ~ = 0, 

rfft 
or dT cos- +Fds = 0, 

Nds-2Tsm—-dTsm- = 0, 
2 2 

where dd is the angle between the two tensile forces which 
act at the ends of the element. The negative sign in F (—ds) 
indicates the fact that F and ds are measured in opposite di- 
rections. But since the cord is supposed to be perfectly 
flexible the tensile forces are tangent to the surface of con- 
tact. Therefore is the angle between the tangents, and 
consequently the angle between the normals, at the ends of 
the element. As an angle becomes indefinitely small its 
cosine approaches unity and its sine approaches the angle 
itself,* therefore we can make the substitutions 

de , , . de de 

cos — = 1 and sin — = — 

2 2 2 

in the last two equations, and obtain 

dT+Fds=0, (1) 

and Nds-Td6 + idTdd=0. (2) 

Neglecting the differential of the second order in equation 
(2) and then eliminating ds between equations (1) and (2) 
we get 

Y = "N de= ~ M ^' (3) 

where m is the coefficient of friction. Integrating the last 

* See Appendix A. 



70 



ANALYTICAL MECHANICS 



equation and passing from the logarithmic to the exponen- 
tial form, we have 

T = ce~»\ 

where c is the constant of integration. If d is measured from 
the normal to the surface at the point where the right-hand 
side of the cord leaves contact we obtain the initial condition, 
T = To when 0=0, which determines c. Applying this con- 
dition to the last equation we have 

T = T e-» e . (4) 

Discussion. — Equation (4) gives the relation between the values of 
the tensile force at any two points of the cord. It must be observed that 
is measured in the same direction as F; in other words, opposite the 
direction towards which the cord is urged to move. Therefore T or T 
has the larger value according to whether is positive or negative. As 
a concrete example suppose a weight W to be suspended from the right- 
hand end of the cord and to be held in equilibrium by a force F applied at 
the left-hand end. If F is just large enough to prevent W from falling 
then the cord will be on the point of moving to the right, therefore is 
measured in the counter-clockwise T 
direction and is positive. In this 
case 

F = We' 116 . 

In case F is just large enough to start 
W to move up, then is measured in 
the clockwise direction and is nega- 
tive. Therefore ^ 

F = We>*. 




The value of T drops very rapidly 
with the increase of 0. This fact 
is made clear by drawing the graph 
of equation (4), Fig. 45. The graph 
may be constructed easily by making use of the half-value period of the 
curve. If P denotes the period, then, by definition, the ordinate is reduced 
to one-half its value every time P is added to 0.* We have therefore , 

±T=T e-» (e+P) . 

* The difference between this definition of P and the one given in the pre- 
ceding section is accounted for by the difference in the signs of the exponents 
in equation (4) and in equation (14) of the preceding section. 



EQUILIBRIUM OF FLEXIBLE CORDS 71 



(5) 



Dividing equation (4) by the last equation we get 

or P=ilog e 2 

= ^. 

Thus if 6 = nP, then by equations (4) and (5) 

T=|- (6) 

Therefore taking 0.53 for hemp rope on oak and 6 = 2 w, we obtain 
n = 4.76 and 2 n = 27.3. Hence in this case T is 27.3 times as great 
as T. 

APPLICATION TO BELTS. 

The tensile force on one side of a belt which transmits 
power is greater than that on the other side. The relation 
between the tensile forces on the two sides of the belt is 
given by equation (4) . Thus if T x denotes the tensile force 
on the driving side and T 2 that on the slack side, then 

T 2 = T&- 1 * or 7\ = T2&*. (40 

The difference between T\ and T 2 is the effective force which 
drives the pulley. Denoting the effective force by F, we 
have 

F = Tx - T 2 

= Tl (l- e~^) (7) 

= !T 2 (e^-l). ) 

We have neglected the cross-section of the cord in the 
solution of the foregoing problem. Therefore the results 
which we have obtained are applicable to actual problems 
only when the cross-section of the cord is negligible com- 
pared with that of the solid with which it is in contact. 

PROBLEMS. 
1. A weight of 5 tons is to be raised from the hold of a ship by means 
of a rope which takes 3| turns around the drum of a steam windlass. If 
(jl = 0.25 what force must a man exert at the other end of the rope? 



72 ANALYTICAL MECHANICS 

2. By pulling with a force of 200 pounds a man just keeps from surg- 
ing a rope, which takes 2.5 turns around a post. Find the tensile force 
at the other end of the rope, /x = 0.2. 

3. A weight W is suspended by a rope which makes li turns around 
a clamped pulley and goes to the hand of a workman. If /x = 0.2, find 
the force the man has to apply in order (a) to support the weight, (b) to 
raise it. 

4. Two men, who can pull 250 pounds each, can support a weight by 
means of a rope which takes 2 turns around a post. On the other hand, 
one of the men can support it alone if the rope makes 2.5 turns. Find 
the weight. 

5. In order to prevent surging a sailor has to exert a force of 150 
pounds at the end of a hawser, which is used to keep the stern of a boat 
at rest while the bow is being turned by the engines. Find the pull 
exerted by the boat upon the hawser under the following conditions : 

[Hint. — Make use of equations (5) and (6).] 

(a) = | 5 M = 0.2. 

(b) = |, M = 0.5. 

(c) = |, M = 0.5. 

(d) d = 7T, m = 0.4. 

(e) d = ^f, M = 0.3. 

(f) = ^, M = 0.2. 

6. A belt has to transmit an effective force of 500 pounds. Find the 
tensile force on both sides of the belt, under the following conditions : 

(a) = 135°, m = 0.5. (e) = 165°, ix = 0.2. 

(b) 6 = 135°, /* = 0.4. (f) 6 = 180°, n = 0.3. 

(c) = 150°, n = 0.3. (g) d = 180°, m = 0.5. 

(d) = 165°, ju = 0.5. (h) = 195° 7 /* = 0.4. 

7. In the preceding problem find the width of the belt, supposing 
the permissible safe tensile force to be 50 pounds per inch of its width. 



(g) B = 2ir, 


IX = 0.1. 


(h) e = ^f, 


ix = 0.4. 


(i) e = 5 f, 


ix = 0.5. 


(i) e = 3w, 


ix = 0.3. 


(k) = ^ 


ix = 0.4. 


(i) e = 7 -f, 


ix = 0.5. 



F£8 J 1913 



c^ 



LBAp'13 



ANALYTICAL MECHANICS 

FOR STUDENTS OF PHYSICS AND 
ENGINEERING 



BY 



HAROUTUNE M. DADOUBIAN, M.A, Ph.D. 

Instructor of Physics in the Sheffield Scientific School 
of Yale University 




NEW YORK 

D. VAN NOSTRAND COMPANY 

25 Park Place 

1913 












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